Problem: The equation of a circle $C$ is $x^2+y^2+8x+6y+24 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+8x) + (y^2+6y) = -24$ $(x^2+8x+16) + (y^2+6y+9) = -24 + 16 + 9$ $(x+4)^{2} + (y+3)^{2} = 1 = 1^2$ Thus, $(h, k) = (-4, -3)$ and $r = 1$.